Maths 4 Marks

The equation of the family of curves is

$\text{x}^2+\text{y}^2=\text{ax}^3\ ...(1)$

where a is a parameter.

As this equation has only one arbitrary constant, we shall get a differential equation of first order.

Differentiating (1) with respect to x, we get

$2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}=3\text{ax}^2$

$\Rightarrow2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}=3\Big(\frac{\text{x}^2+\text{y}^2}{\text{x}^3}\Big)\text{x}^2$

$\Rightarrow2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}=3\frac{\text{x}^2+\text{y}^2}{\text{x}^3}$

$\Rightarrow2\text{x}^2+2\text{xy}​​=3\text{x}^2+3\text{y}^2$

$2\text{xy}​​=\text{x}^2+3\text{y}^2$

It is the required differential equation.

$(1+\text{x}^2)\text{dy}=\text{xy dx}$

$\Rightarrow\frac{1}{\text{y}}\text{dy}=\frac{\text{x}}{1+\text{x}^2}\ \text{dx}$

Integrating both sides, we get

$\int\frac{1}{\text{y}}\text{dy}=\int\frac{\text{x}}{1+\text{x}^2}\ \text{dx}$

Substituting 1+ x² = t, we get

$2\text{x dx}=\text{dt}$

$\therefore\int\frac{1}{\text{y}}\text{dy}=\frac{1}{2}\int\frac{1}{\text{t}}\text{dt}$

$\Rightarrow\log|\text{y}|=\frac{1}{2}\log|\text{t}|+\log\text{C}$

$\Rightarrow\log|\text{y}|=\frac{1}{2}\log|1+\text{x}^2|+\log\text{C}$

$\Rightarrow\log|\text{y}|=\log\Big[\text{C}\sqrt{1+\text{x}^2}\Big]$

$\Rightarrow\text{y}=\text{C}\sqrt{1+\text{x}^2}$

Hence, $\text{y}=\text{C}\sqrt{1+\text{x}^2}$ is the required solution. 

The equation of the family of circles that pass through the origin (0, 0) and whose centres lie on the x-axis is given by

$(\text{x}-\text{a})^2+\text{y}^2=\text{a}^2\ ...(1)$

 where a are arbitrary constants.

As this equation has only one arbitrary constant, we shall get a first order differential equation.

Differentiating (1) with respect to x, we get

$2(\text{x}-\text{a})+2\text{y}\frac{\text{dy}}{\text{dx}}=0$

$\Rightarrow\text{x}-\text{a}+\text{y}\frac{\text{dy}}{\text{dx}}=0$

$\Rightarrow\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}=\text{a}$

Substituting the value of a in equation (2), we get

$\Big(\text{x}-\text{x}-\text{y}​​\frac{\text{dy}}{\text{dx}}\Big)^2+\text{y}^2=\Big(\text{x}+\text{y}​​\frac{\text{dy}}{\text{dx}}\Big)^2$

$\Rightarrow\text{y}^2\Big(​​\frac{\text{dy}}{\text{dx}}\Big)^2+\text{y}^2=\text{x}^2+2\text{xy}​​\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big(​​\frac{\text{dy}}{\text{dx}}\Big)^2$

$\Rightarrow2\text{xy}​​\frac{\text{dy}}{\text{dx}}+\text{x}^2=\text{y}^2$

It is the required differential equation.

$\frac{\text{dy}}{\text{dx}}=\frac{\text{x e }^\text{x}\log\text{x}+\text{e}^\text{x}}{\text{x}\cos\text{y}}$

$\Rightarrow\text{x}\cos\text{y dy}=(\text{x e}^\text{x}\log\text{x}+\text{e}^\text{x})\ \text{dx}$

$\Rightarrow\cos\text{y dy}=\Big(\text{e}^\text{x}\log\text{x}+\frac{1}{\text{x}}\text{e}^\text{x}\Big)\ \text{dx}$

Integrating both sides, we get

$\int\cos\text{y dy}=\int\Big(\text{e}^\text{x}\log\text{x}+\frac{1}{\text{x}}\text{e}^\text{x}\Big)\text{dx}$

$\Rightarrow\sin\text{y}=\log\text{x}\int\text{e} ^\text{x}\text{dx}-\int\frac{1}{\text{x}}\text{e}^\text{x}\text{dx}+\int\frac{1}{\text{x}}\text{e}^\text{x}\text{dx}$

$\Rightarrow\sin\text{y}=\text{e}^\text{x}\log\text{x}+\text{C}$

Hence, $\sin\text{y}=\text{e}^\text{x}\log\text{x}+\text{C}$ is the required solution.

The equation of the family of curves is

$\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1$

where a is a parameter.

As this equation has only one arbitrary constant, we shall get a differential equation of first order.

Differentiating (1) with respect to x, we get

$\frac{2\text{x}}{\text{a}^2}-\frac{2\text{y}}{\text{b}^2}\frac{\text{dy}}{\text{dx}}=0\ ...(2)$

Differentiating (2) with respect to x, we get

$\frac{2}{\text{a}^2}-\frac{2}{\text{b}^2}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-\frac{2\text{y}}{\text{b}^2}\frac{\text{d}^2\text{y}}{\text{dx}^2}=0$

$\Rightarrow\frac{2}{\text{a}^2}=\frac{2}{\text{b}^2}\Big[\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big]$

$\Rightarrow\frac{\text{b}^2}{\text{a}^2}=\Big[\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big]\ ...(3)$

Now, from (2), we get

$\frac{2\text{x}}{\text{a}^2}=\frac{2\text{y}}{\text{b}^2}\frac{\text{dy}}{\text{dx}}$

$\Rightarrow\frac{\text{b}^2}{\text{a}^2}=\frac{\text{y}^2}{\text{x}}\frac{\text{dy}}{\text{dx}}$

From (3), (4), we get

$\frac{\text{y}}{\text{x}}\frac{\text{dy}}{\text{dx}}=\Big[\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big]$

$\Rightarrow\text{x}\Big[\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big]=\text{y}\frac{\text{dy}}{\text{dx}}$

It is the required differential equation.

The equation of the family of hyperbolas having the centre at the origin and foci on the x-axis is

$\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1\ ...(1)$

where a and b are parameters. As this equation contains two parameters, we shall get a second-order differential equation.

Differentiating equation (1) with respect to x, we get

$\frac{2\text{x}}{\text{a}^2}-\frac{2\text{y}}{\text{b}^2}\frac{\text{dy}}{\text{dx}}=0\ ...(2)$

Differentiating equation (2) with respect to x, we get

$\frac{2}{\text{a}^2}-\frac{2}{\text{b}^2}\Big[\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big]=0$

$\Rightarrow\frac{1}{\text{a}^2}=\frac{1}{\text{b}^2}\Big[\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big]$

$\Rightarrow\frac{\text{b}^2}{\text{a}^2}=\Big[\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big]$

Now, from equation (2), we get

$\frac{2\text{x}}{\text{a}^2}=\frac{2\text{y}}{\text{b}^2}\frac{\text{dy}}{\text{dx}}$

$\Rightarrow\frac{\text{b}^2}{\text{a}^2}=\frac{\text{y}}{\text{x}}\frac{\text{dy}}{\text{dx}}\ ...(4)$

From (3) and (4), we get

$\frac{\text{y}}{\text{x}}\frac{\text{dy}}{\text{dx}}=\Big[\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big]$

$\Rightarrow\text{y}\frac{\text{dy}}{\text{dx}}=\text{xy}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$

$\Rightarrow\text{xy}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-\text{y}\frac{\text{dy}}{\text{dx}}=0$

It is the required differential equation.

We have,

$\text{y}=4\text{ax}\ ...(1)$

Differentiating both sides of equation (1) with respect to 3, we get

$2\text{y}\frac{\text{dy}}{\text{dx}}=4\text{a}$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2\text{a}}{\text{y}}\ ...(2)$

now, differentiating both sided of (1) with respect to y, we get

$2\text{y}=4\text{a}\frac{\text{dy}}{\text{dx}}$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{2\text{a}}\ ...(3)$

$\therefore\text{x}\frac{\text{dy}}{\text{dx}}+\text{a}\frac{\text{dy}}{\text{dx}}=\text{x}\Big(\frac{2\text{a}}{\text{y}}\Big)+\text{a}\Big(\frac{\text{y}}{2\text{a}}\Big)$

$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}+\text{a}\frac{\text{dy}}{\text{dx}}=\frac{2\text{ax}}{\text{y}}+\frac{\text{y}}{2}$

$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}+\text{a}\frac{\text{dy}}{\text{dx}}=\frac{\text{y}^2}{2\text{y}}+\frac{\text{y}}{2}$

$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}+\text{a}\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{2}+\frac{\text{y}}{2}$

$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}+\text{a}\frac{\text{dy}}{\text{dx}}=\text{y}$

$\Rightarrow\text{y}=\text{x}\frac{\text{dy}}{\text{dx}}+\text{a}\frac{\text{dy}}{\text{dx}}$

Hence, the given function is the solution to the given differential equation.

The equation of the family of curves is

y2 - 2ay + 2+ = a² ...(1)

 where a is a parameter.

This equation contains only one arbitrary constant, so we shall get a differential equation of first order.

Differentiating equation (1) with respect to x, we get

$2\text{y}\frac{\text{dy}}{\text{dx}}-2\text{a}\frac{\text{dy}}{\text{dx}}+2\text{x}=0$

$\Rightarrow2\text{y}\frac{\text{dy}}{\text{dx}}+2\text{x}=2\text{a}\frac{\text{dy}}{\text{dx}}$

$\Rightarrow\text{y}+\frac{\text{x}}{\frac{\text{dy}}{\text{dx}}}=\text{a}$

Substituting the value of a in equation (2), we get

$\text{y}^2-2\Bigg(\text{y}+\frac{\text{x}}{\frac{\text{dy}}{\text{dx}}}\Bigg)\text{y}+\text{x}^2=\Bigg(\text{y}+\frac{\text{x}}{\frac{\text{dy}}{\text{dx}}}\Bigg)^2$

$\Rightarrow\frac{\text{y}^2\frac{\text{dy}}{\text{dx}}-2\Big(\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}\Big)\text{y}+\text{x}^2\frac{\text{dy}}{\text{dx}}}{\frac{\text{dy}}{\text{dx}}}=\frac{\Big(\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}\Big)}{\Big(\frac{\text{dy}}{\text{dx}}\Big)^2}$

$\Rightarrow\text{y}^2\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-2\text{y}^2\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-2\text{xy}\Big(\frac{\text{dy}}{\text{dx}}\Big)+\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\\=\text{y}^2\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+2\text{xy}\Big(\frac{\text{dy}}{\text{dx}}\Big)+\text{x}^2$

$\Rightarrow(\text{x}^2-2\text{y}^2)\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-4\text{xy}\Big(\frac{\text{dy}}{\text{dx}}\Big)-\text{x}^2=0$

It is the required differential equation.

The equation of the family of parabolas with latus rectum 4a and axis parallel to the x-axis is given by

$(\text{y}-\beta)^2=4\text{a}(\text{x}-\text{a})\ ...(1)$

where a and Bare two arbitrary constants.

As this equation has two arbitrary constants, we shall get second order differential equation.

Differentiating equation (1) with respect to x, we get

$2(\text{y}-\beta)\frac{\text{dy}}{\text{dx}}=4\text{a}\ ...(2)$

 Differentiating equation (2) with respect to x, we get

$(\text{y}-\beta)\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\frac{\text{dy}}{\text{dx}}\Big)=0$

Now, from equation (2) we get,

$(\text{y}-\beta)=\frac{4\text{a}}{\frac{\text{dy}}{\text{dx}}}$

From (3) and (4), we get

$\frac{2\text{a}}{\frac{\text{dy}}{\text{dx}}}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=0$

$\Rightarrow2\text{a}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^3=0$

It is the required differential equation.

$\text{x}\cos^2\text{y dx}=\text{y}\cos^2\text{x dx}$

$\Rightarrow\frac{\text{x}}{\cos^2\text{x}}\ \text{dx}=\frac{\text{y}}{\cos^2\text{y}}\ \text{dy}$

$\Rightarrow\text{x}\sec^2\text{x dx}=\text{y}\sec^2\text{y dy}$

Integrating both sides, we get

$\int\text{x}\sec^2\text{x dx}=\int\text{y}\sec^2\text{y dy}$

$\Rightarrow\text{x}\int\sec\text{x dx}-\int\Big\{\frac{\text{d}}{\text{dx}}(\text{x})\int\sec^2\text{x dx}\Big\}\text{dx}\\=\text{y}\int\sec^2\text{y dy}-\int\Big\{\frac{\text{d}}{\text{dy}}(\text{y})\int\sec^2\text{y dy}\Big\}\text{dy}$

$\Rightarrow\text{x}\tan\text{x}-\int\int\tan\text{x dx}=\text{y}\tan\text{y}-\int\tan\text{y dy}$

$\Rightarrow\text{x}\tan\text{x}-\log|\sec\text{x}|=\text{y}\tan\text{y}-\log|\sec\text{y}|+\text{C}$

$\Rightarrow\text{x}\tan\text{x}-\text{y}\tan\text{y}=\log|\sec\text{x}|-\log|\sec\text{y}|+\text{C}$

Hence, $\text{x}\tan\text{x}-\text{y}\tan\text{y}=\log|\sec\text{x}|-\log|\sec\text{y}|+\text{C}$ is the required solution.

Portion of the x-axis cut off between the origin and tangent at a point  $=\text{x}-\text{y}\frac{\text{dx}}{\text{dy}}=\text{OT}$

It is given, $\text{OT}=2\text{x}$ 

$\therefore \ \text{x}-\text{y}\frac{\text{dx}}{\text{dy}}=2\text{x}$

$-\text{x}=\text{y}\frac{\text{dx}}{\text{dy}}$

$-\int\frac{\text{dx}}{\text{dy}}=\int\frac{\text{dy}}{\text{y}}$

$\therefore\ \text{xy}=\text{k}$

Since the curve passes through the point (1, 2)

at  $\text{x}=1, \text{y}=2$

$\therefore \text{k}=2$

$\therefore \text{xy}=2$

We know that, equation of a dide with centre at (h, k) and radius r is given by,

(x - h²) + (y - k)² = r² ...(1)

Here, entre lies, on y-axis, so h = 0

x² + (y - k)² = r² ...(2)

Also given that circle is passing through origin, so

0 + k² = r²

k² = r²

So, equation (2) becomes

x² +  (y - k)² =  k²

x² + y² - 2yk = 0

2yk = x² + y²

$\text{k}=\frac{\text{x}^2+\text{y}^2}{2\text{y}}$

Differentiating with respect to x,

$0=\frac{\Big(2\text{y}2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}\Big)-(\text{x}^2+\text{y}^2)2\frac{\text{dy}}{\text{dx}}}{(2\text{y})^2}$

$0=4\text{xy}+4\text{y}^2\frac{\text{dy}}{\text{dx}}-2\text{x}^2\frac{\text{dy}}{\text{dx}}-2\text{y}^2\frac{\text{dy}}{\text{dx}}$

$0=2\text{y}^2\frac{\text{dy}}{\text{dx}}-2\text{x}^2\frac{\text{dy}}{\text{dx}}+4\text{xy}$

$\text{x}^2\frac{\text{dy}}{\text{dx}}-\text{y}^2\frac{\text{dy}}{\text{dx}}=2\text{xy}$

$(\text{x}^2-\text{y}^2)\frac{\text{dy}}{\text{dx}}=2\text{xy}$

The equation of the family of curves is

$(\text{x}-\text{a})^2+(\text{y}-\text{b})^2=\text{r}^2\ ...(1)$

 where a and b is a parameter.

This equation contains only one arbitrary constant, so we shall get a differential equation of first order.

Differentiating equation (1) with respect to x, we get

$2(\text{x}-\text{a})+2(\text{y}-\text{b})\frac{\text{dy}}{\text{dx}}=0\ ...(2)$

Differentiating equation (1) with respect to x, we get

$2+2\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+2(\text{y}-\text{b})\frac{\text{d}^2\text{y}}{\text{dx}^2}=0$

$\Rightarrow1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+(\text{y}-\text{b})\frac{\text{d}^2\text{y}}{\text{dx}^2}=0$

$\Rightarrow(\text{y}-\text{b})=\frac{1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2}{\frac{\text{d}^2\text{y}}{\text{dx}^2}}$

From (2) and (3), we get

$(\text{x}-\text{a})-\frac{1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2}{\frac{\text{d}^2\text{y}}{\text{dx}^2}}\frac{\text{dy}}{\text{dx}}=0\Rightarrow(\text{x}-\text{a})=\frac{\frac{\text{dy}}{\text{dx}}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^3}{\frac{\text{d}^2\text{y}}{\text{dx}^2}}\ ...(4)$

From (1) and (3), we get

$\frac{\Big[\frac{\text{dy}}{\text{dz}}+\Big(\frac{\text{dy}}{\text{dz}}\Big)^2\Big]^2}{\Big(\frac{\text{d}^2\text{y}^2}{\text{dz}^2}\Big)^2}+\frac{\Big[1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big]^2}{\Big(\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)^2}=\text{r}^2$

$\Rightarrow\frac{\Bigg[\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+2\Big(\frac{\text{dy}}{\text{dx}}\Big)^4+\Big(\frac{\text{dy}}{\text{dx}}\Big)^6\Bigg]+\Bigg[1+2\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+\Big(\frac{\text{dy}}{\text{dx}}\Big)^4\Bigg]}{\Big(\frac{\text{d}^2\text{y}}{\text{dx}}\Big)^2}$

$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+2\Big(\frac{\text{dy}}{\text{dx}}\Big)^4+\Big(\frac{\text{dy}}{\text{dx}}\Big)^6+1+2\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+\Big(\frac{\text{dy}}{\text{dx}}\Big)^4=\text{r}^2\Big(\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)$

$\Rightarrow1+3\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+3\Big(\frac{\text{dy}}{\text{dx}}\Big)^4+\Big(\frac{\text{dy}}{\text{dx}}\Big)^6=\text{r}^2\Big(\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)$

$\Rightarrow\Bigg[1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Bigg]^3=\text{r}^3\Big(\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)^2$

It is the required differential equation.

We have,

$\text{y}=\text{A}\cos2\text{x}+\text{B}\sin2\text{x}\ ...(1)$

Differentiating both sides of equation (1) with respect to 3, we get

$\frac{\text{dy}}{\text{dx}}=-2\text{A}\sin2\text{x}-\text{B}\cos2\text{x}\ ...(2)$

Differentiating both sides of equation (2) with respect to 3, we get

$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-4\text{A}\cos2\text{x}+4\text{B}\sin2\text{x}$

$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-4(\text{A}\cos2\text{x}+4\text{B}\sin2\text{x})$

$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-4\text{y}$

$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}+4\text{y}=0$

Hence, the given function is the solution to the given differential equation.

We have,

$\text{y}=\text{ae}^{2\text{x}}+\text{be}^{-\text{x}}\ ...(1)$

Differentiating both sides of equation (1) with respect to 3, we get

$\frac{\text{dy}}{\text{dx}}=\text{ae}^{2\text{x}}+\text{be}^{-\text{x}}\ ...(2)$

Differentiating both sides of equation (2) with respect to 3, we get

$\frac{\text{d}^2\text{y}}{\text{dx}^2}=4\text{ae}^{2\text{x}}+\text{be}^{-\text{x}}$

$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\text{ae}^{2\text{x}}-\text{be}^{-\text{x}}+2\text{ae}^{2\text{x}}+2\text{be}^{-\text{x}}$

$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\big(2\text{ae}^{2\text{x}}-\text{be}^{-\text{x}}\big)+2\big(\text{ae}^{2\text{x}}+2\text{be}^{-\text{x}}\big)$

$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{dy}}{\text{dx}}+2\text{y}$

$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}-\frac{\text{dy}}{\text{dx}}-2\text{y}=0$

Hence, the given function is the solution to the given differential equation.

we know that the equation of said family of ellopsis is

$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1\ ...(1)$

Differentiating (1) w..r.t.x, we get

$\frac{2\text{x}}{\text{a}^2}+\frac{2\text{y}}{\text{b}^2}.\frac{\text{dy}}{\text{dx}}=0$

$\Rightarrow\frac{\text{y}}{\text{x}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{-\text{b}^2}{\text{a}^2}\ ...(2)$

Differentiating (2) w..r.t.x, we get

$\frac{\text{y}}{\text{x}}\Big(\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)+\bigg(\frac{\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}}{\text{x}^2}\bigg)\frac{\text{dy}}{\text{dx}}=0$

$\Rightarrow\text{xy}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-\text{y}\frac{\text{dy}}{\text{dx}}$

Which is the required difeerential equation.

$\frac{\text{dy}}{\text{dx}}=\frac{\text{x + y}}{\text{x}-\text{y}}$

This is a homogeneous differential equation.

Put y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$, we get

$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{{\text{x}}+\text{vx}}{\text{x}-\text{vx}}$

$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{1+\text{v}}{1-\text{v}}$

$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+\text{v}}{1-\text{v}}-\text{v}$

$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+\text{v}^2}{1-\text{v}}$

$\frac{1-\text{v}}{1+\text{v}^2}\text{dv}=\frac{1}{\text{x}}\text{dx}$

Integrating both sides, we get

$\int\frac{1-\text{v}}{1+\text{v}^2}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$

$\Rightarrow\ \int\frac{1}{1+\text{v}^2}\text{dv}-\int\frac{\text{v}}{1+\text{v}^2}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$

$\Rightarrow\ \int\frac{1}{1+\text{v}^2}\text{dv}-\frac{1}2\int\frac{2\text{v}}{1+\text{v}^2}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$

$\tan^{-1}\text{v}-\frac{1}2\log\big|1+\text{v}^2\big|=\log|\text{x}|+\text{C}$

Putting $\text{v}=\frac{\text{y}}{\text{x}}$, we get

$\Rightarrow\ \tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)-\frac{1}2\log\Big|1+\frac{\text{y}^2}{\text{x}^2}\Big|=\log|\text{x}|+\text{C}$

$\Rightarrow\ \tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)=\frac{1}2\log\Big|1+\frac{\text{y}^2}{\text{x}^2}\Big|+\log|\text{x}|+\text{C}$

$\Rightarrow\ \tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)=\frac{1}2\log\Big|\frac{\text{x}^2+\text{y}^2}{\text{x}^2}\Big|+\log|\text{x}|+\text{C}$

$\Rightarrow\ \tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)=\frac{1}2\log\Big|\text{x}^2+\text{y}^2\Big|-\frac{1}2\log|\text{x}^2|+\log|\text{x}|+\text{C}$

$\Rightarrow\ \tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)=\frac{1}2\log\Big|\text{x}^2+\text{y}^2\Big|-\log|\text{x}|+\log|\text{x}|+\text{C}$

$\Rightarrow\ \tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)=\frac{1}2\log\Big|\text{x}^2+\text{y}^2\Big|+\text{C}$

Hence, $\tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)=\frac{1}2\log\Big|\text{x}^2+\text{y}^2\Big|+\text{C}$ is the required solution.

It is given that the distance between the foot of ordinate of point of contanct (A) and the point of intersection of tangent with x-axis (T) = 2x

Coordinate of  $\text{T}=\big(\text{x}-\text{y}\frac{\text{dx}}{\text{dy}}, 0\big)$

$\text{AT}=\Big[\text{x}-\Big(\text{x}-\text{y}\frac{\text{dx}}{\text{dy}}, 0\Big)\Big]=2\text{x}$

Equation of tangent, 

$\text{y}-\text{y}=\frac{\text{dy}}{\text{dx}}(\text{x}-\text{x})$

$\Rightarrow\text{y}-\text{0}=\frac{\text{dy}}{\text{dx}}\Big(\text{x}-\big(\text{x}-\text{y}\frac{\text{dx}}{\text{dy}}\big)\Big)$

$\Rightarrow \text{y}\frac{\text{dx}}{\text{dy}}=2\text{x}$

$\Rightarrow \int\frac{\text{dx}}{\text{x}}=2\int\frac{\text{dy}}{\text{y}}$

$\Rightarrow \log\text{x}=\log\text{y}^{2}+\log\text{C}$

$\text{x}=\text{Cy}^{2}$

As the circle passes through (1, 2)

$1=\text{C}\times2^{2}$

$\Rightarrow \text{C}=\frac{1}{4}$

$\Rightarrow 4\text{x}=\text{y}^{2}$

We have,

$\text{y}=\frac{\text{a}}{\text{x}}+\text{b}\ ...(1)$

Differentiating both sides of equation (1) with respect to 3, we get

$\frac{\text{dy}}{\text{dx}}=-\frac{\text{a}}{\text{x}^2}\ ...(2)$

Differentiating both sides of equation (2) with respect to 3, we get

$\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\frac{\text{a}}{\text{x}^3}$

$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{2}{\text{x}}\Big(-\frac{\text{a}}{\text{x}^2}\Big)$

$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{2}{\text{x}}\Big(\frac{\text{dy}}{\text{dx}}\Big)$

$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}+\frac{2}{\text{x}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=0$

Hence, the given function is the solution to the given differential equation.

It is homogeneous equation

Put y = vx

$\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$

So,

$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}^2\text{x}^2-\text{x}^2}{2\text{xvx}}$

$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}^2-1}{2\text{v}}-\frac{\text{v}}1$

$=\frac{\text{v}^2-1-2\text{v}^2}{2\text{v}}$

$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{-1-\text{v}^2}{2\text{v}}$

$\int\frac{2\text{v}}{1+\text{v}^2}\text{dv}=-\int\frac{\text{dx}}{\text{x}}$

$\log\big|1+\text{v}^2\big|=-\log|\text{x}|+\log|\text{C}|$

$1+\text{v}^2=\frac{\text{C}}{\text{x}}$

$1+\frac{\text{y}^2}{\text{x}^2}=\frac{\text{C}}{\text{x}}$

$\text{x}^2+\text{y}^2=\text{Cx}$

$\text{C}'(\text{x})\text{dx}=(2+0.15\text{x})\text{dx}$

$\int\text{C}'(\text{x})\text{dx}=\int2\text{dx}+0.15\int\text{x dx}$

$\text{C}(\text{x})=2\text{x}+0.15\frac{\text{x}^2}{2}+\text{C}\ ...(1)$

Put x = 0, c(x) = 100

100 = 2(0) + 0 + c

100 = c

Put c = 100 in equation 1

$\text{c}(\text{x})=2\text{x}+(0.15)\frac{\text{x}^2}{2}+100$ 

We have,

$\text{y}=\text{A}\cos\text{x}+\text{B}\sin\text{x}\ ...(1)$

Differentiating both sides of equation (1) with respect to 3, we get

$\frac{\text{dy}}{\text{dx}}=\text{A}\cos\text{x}+\text{B}\sin\text{x}\ ...(2)$

Differentiating both sides of equation (2) with respect to 3, we get

$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{A}\cos\text{x}-\text{B}\sin\text{x}$

$\frac{\text{d}^2\text{y}}{\text{dx}^2}=(\text{A}\cos\text{x}+\text{B}\sin\text{x})$

$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\text{y}$

$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{y}=0$

Hence, the given function is the solution to the given differential equation.

$\text{dy}=\text{x}^5\tan^{-1}(\text{x}^3)\text{dx}$

$\int\text{dy}=\int\text{x}^5\tan^{-1}(\text{x}^3)\text{dx}$

put $\text{x}^3=\text{t}$

$\Rightarrow3\text{x}^2\text{dx}=\text{dt}$

$\Rightarrow\text{x}^2\text{dx}=\frac{\text{dt}}{3}$

So,

$\int\text{dy}=\frac{1}{3}\Big[\tan^{-1}\text{t}\int\text{t dt}=\int\Big(\frac{1}{1+\text{t}^2}\Big)\times\int\text{t dx}\Big)\Big]\text{dt}+\text{C}$

Using integration by parts

$\text{y}=\frac{1}{3}\Big[\frac{\text{t}^2}{2}+\tan^{-1}-\int\frac{\text{t}^2}{2(\text{t}^2+1)}\text{dt}\Big]+\text{C}$

$=\frac{1}{6}\text{t}^2\tan^{-1}\text{t}-\frac{1}{6}\int\Big(\frac{\text{t}^2}{\text{t}^2+1}\Big)\text{dt}+\text{C}$

$\text{y}=\frac{1}{6}\text{t}^2\tan^{-1}\text{t}-\frac{1}{6}\int\Big(1-\frac{1}{\text{t}^2+1}\Big)\text{dt}+\text{C}$

$=\frac{1}{6}\text{t}^2\tan^{-1}\text{t}-\frac{1}{6}\text{t}+\frac{1}{6}\tan^{-1}\text{t}+\text{C}$

$\text{y}=\frac{1}{6}(\text{t}^2+1)\tan^{-1}\text{t}-\frac{1}{6}\text{t}+\text{C}$

$\text{y}=\frac{1}{6}[(\text{t}^2+1)\tan^{-1}\text{t}-\text{t}]+\text{C}$

So,

$\text{y}=\frac{1}{6}[(\text{x}^6+1)\tan^{-1}(\text{x}^3)-\text{x}^3]+\text{C}$

Let A be the surface area of rain drain, v be its volume, and r be the radius of rain drop.

Given:

$\frac{\text{dV}}{\text{dt}}=\text{A}$

$\frac{\text{dV}}{\text{dt}}=-\text{KA}$ [negative because V decrease with increase in t]

Where K is the constant of proportionality.

So,

$\frac{\text{d}}{\text{dt}}\Big(\frac{4\pi}{3}\text{r}^3\Big)=-\text{K}(4\pi\text{r}^2)$

$4\pi\text{r}^2\frac{\text{dr}}{\text{dt}}=-\text{K}(4\pi\text{r}^2)$

$\frac{\text{dr}}{\text{dt}}=-\text{K}$

$(\sin\text{x}+\cos\text{x})\text{dy}+(\cos\text{x}+\sin\text{x})\text{dx}=0$

$\Rightarrow{\text{dy}}=-\Big(\frac{\cos\text{x}-\sin\text{x}}{\sin\text{x}-\cos\text{x}}\Big)\text{dx}$

Integrating both sides, we get

$\int{\text{dy}}=-\int\Big(\frac{\cos\text{x}-\sin\text{x}}{\sin\text{x}-\cos\text{x}}\Big)\text{dx}$

$\Rightarrow\text{y}=-\int\Big(\frac{\cos\text{x}-\sin\text{x}}{\sin\text{x}-\cos\text{x}}\Big)\text{dx}$

Putting $\sin\text{x}+\cos\text{x}=\text{t}$

$\Rightarrow(\cos\text{x}-\sin\text{x})\text{dx}=\text{dt}$

$\therefore\text{y}=-\int\frac{\text{dt}}{\text{t}}$

$\Rightarrow\text{y}=-\log|\text{t}|+\text{C}$

$\Rightarrow\text{y}=-\log|\sin\text{x}+\cos\text{x}|+\text{C}$

$\Rightarrow\text{y}=+\log|\sin\text{x}+\cos\text{x}|=\text{C}$

Hence, $\text{y}=+\log|\sin\text{x}+\cos\text{x}|=\text{C}$ is the solution to the given differential equation. 

The given equation is

$\text{y}=2(\text{x}^2-1)+\text{ce}^{-\text{x}^{2}} ...(1)$

Where c is a parameter.

As this equation has one arbitrary constant, we shall get a differential equation of first order.

Differentiating equation (1) with respect to x, we get

$\frac{\text{dy}}{\text{dx}}=2(2\text{x})+\text{ce}^{-\text{x}^{2}}(-2\text{x})$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=4\text{x}-2\text{xce}^{-\text{x}^{2}}$

From (1) and (2), we get

$\frac{\text{dy}}{\text{dx}}=4\text{x}-2\text{x}[\text{y}-2\text{x}^2+2]$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=4\text{x}-2\text{xy}+4\text{x}^3-4\text{x}$

$\Rightarrow\frac{\text{dy}}{\text{dx}}+2\text{xy}=4\text{x}^3$

Hence, $\text{y}=2(\text{x}^2-1)+\text{ce}^{-\text{x}^{2}}$ is the solution to the differential equation $\frac{\text{dy}}{\text{dx}}+2\text{xy}=4\text{x}^3.$

Solution of the equation is given by,

$\text{y}\times(\text{I.F.})=\int\text{Q}\times(\text{I.F.})\text{dx + C}$

Let C denote the family of circles in the second qwdrant and touching the coordinate axes.

Let (-a, a) be the coordinate of the centre of any member of this family,

Equation representing the family C is

$(\text{x}+\text{a})^2+(\text{y}-\text{a})^2=\text{a}^2\ ...(1)$

$\text{x}^2+\text{y}^2+2\text{ax}-2\text{ay}+\text{a}^2=0\ ...(2)$

Differentiating eqn (i) w.r.t.x, we get

$2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}+2\text{a}-2\text{a}\frac{\text{dy}}{\text{dx}}=0$

$\Rightarrow\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}=\text{a}\Big(\frac{\text{dy}}{\text{dx}}-1\Big)$

$\Rightarrow\text{a}=\frac{\text{x}+\text{yy'}}{\text{y}-1}$

Substituting the value of a in (ii), we get

$\Big[\text{x}+\frac{\text{x}+\text{yy}'}{\text{y}-1}\Big]^2+\Big[\text{y}-\frac{\text{x}+\text{yy}'}{\text{y}-1}\Big]^2=\Big[\frac{\text{x}+\text{yy}'}{\text{y}-1}\Big]^2$

$\Rightarrow[\text{xy}-\text{x}+\text{x}+\text{yy'}]^2+[\text{yy}'-\text{y}-\text{x}-\text{yy}']^2=[\text{x}+\text{yy'}]^2$

$\Rightarrow(\text{x}+\text{y})^2\text{y}^2+(\text{x}+\text{y})^2=[\text{x}+\text{yy'}]$

$(\text{x}+\text{y})^2\Big[(\text{y})^2+1\Big]=[\text{x}+\text{yy'}]^2$

which is the differential equation representing the given family of circles.

The equation of the family of curves is

$(2\text{x}-\text{a})^2-\text{y}^2=\text{a}^2 ...(1)$

where a is a parameter.

As this equation has only one parameter, we shall get a differential equation of first order.

Differentiating (1) with respect to x, we get

$8\text{x}-4\text{a}-2\text{y}\frac{\text{dy}}{\text{dx}}=0\ ...(2)$

$\Rightarrow-\text{y}\frac{\text{dy}}{\text{dx}}+4\text{x}=2\text{a}$

Now, from (1), we get

$2\text{a}=\frac{4\text{x}^2-\text{y}^2}{2\text{x}}\ ...(3)$

From (2) and (3), we get

$-\text{y}\frac{\text{dy}}{\text{dx}}+4\text{x}=\frac{4\text{x}^2-\text{y}^2}{2\text{x}}$

$\Rightarrow-2\text{xy}\frac{\text{dy}}{\text{dx}}+8\text{x}^2=4\text{x}^2-\text{y}^2$

$\Rightarrow-2\text{xy}\frac{\text{dy}}{\text{dx}}+4\text{x}^2+\text{y}^2=0$

$\Rightarrow2\frac{\text{dy}}{\text{dx}}=4\text{x}^2+\text{y}^2$

It is the required differential equation.

Multiplying both sides of (1) by e^3x, we get

$\text{x}\frac{\text{dy}}{\text{dx}}+2\text{y}=\text{x}^2\log\text{x}$

$\frac{\text{dy}}{\text{dx}}+\frac{2\text{y}}{\text{x}}=2\log\text{x}$

Comparing with $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q},$ we get

$\text{P}=\frac{2}{\text{x}}$

$\text{Q}=\text{x}\log\text{x}$

$\text{I.F}=\text{e}^{\int\text{Pdx}}$

$=\text{e}^{\int\frac{2\text{}}{\text{x}}}\text{ dx}$

$=\text{e}^{2\log|\text{x}|}$

$=\text{x}^2$

$\text{y}\times\text{I.F.}=\int\text{Q}\times\text{I.F.}\text{ dx}+\text{C}$

$\Rightarrow\text{x}^2\text{y}=\int\text{x}^3\log\text{x dx}+\text{C}$

$\Rightarrow\text{x}^2\text{y}=\log\text{x}\int\text{x}^3\text{dx}-\int\Big[\frac{\text{d}}{\text{dx}}(\log\text{x})\int\text{x}^3\text{dx}\Big]\text{dx}+\text{C}$

$\Rightarrow\text{x}^2\text{y}=\frac{\text{x}^4\log\text{x}}{4}-\int\frac{\text{x}^3}{4}\text{dx}+\text{C}$

$\Rightarrow\text{x}^2\text{y}=\frac{\text{x}^4\log\text{x}}{4}-\frac{\text{x}^3}{16}\text{dx}+\text{C}$

$\Rightarrow\text{y}=\frac{\text{x}^2\log\text{x}}{4}-\frac{\text{x}^2}{16}+\frac{\text{C}}{\text{x}^2}$

$\Rightarrow\text{y}=\frac{\text{x}^2}{16}(4\log\text{x}-1)+\frac{\text{C}}{\text{x}^2}$

The equation of the family of curves is

$(2\text{x}+\text{a})^2+\text{y}^2=\text{a}^2\ ...(1)$

where a is a parameter.

As this equation has only one parameter, we shall get a differential equation of first order.

Differentiating (1) with respect to x, we get

$2(2\text{x}+\text{a})\times2+2\text{y}\frac{\text{dy}}{\text{dx}}=0\ ...(2)$

Now, from (1), we get

$4\text{x}^2+4\text{ax}+\text{a}^2+\text{y}^2=\text{a}^2$

$\Rightarrow4\text{ax}=-\text{y}^2-4\text{x}^2$

$\Rightarrow\text{a}=-\frac{(4\text{x}^2+\text{y}^2)}{4\text{x}}$

putting the value of a in (2), we get

$4\Big(2\text{x}-\frac{4\text{x}^2+\text{y}^2}{4\text{x}}\Big)+2\text{y}\frac{\text{dy}}{\text{dx}}=0$

$\Rightarrow4\Big(\frac{8\text{x}^2-4\text{x}^2-\text{y}^2}{4\text{x}}\Big)+2\text{y}\frac{\text{dy}}{\text{dx}}=0$

$\Rightarrow4\text{x}^2-\text{y}^2+2\text{xy}\frac{\text{dy}}{\text{dx}}=0$

$\Rightarrow\text{y}^2-4\text{x}^2-2\text{xy}\frac{\text{dy}}{\text{dx}}=0$

It is the required differential equation.

The equation of the family of curves is

$(\text{x}-\text{a})^2+2\text{y}^2=\text{a}^2\ ...(1)$

where a is a parameter.

As this equation has only one parameter, we shall get a differential equation of first order.

Differentiating (1) with respect to x, we get

$2\text{x}-2\text{a}+4\text{y}\frac{\text{dy}}{\text{dx}}=0\ ...(2)$

Now, from (1), we get

$2\text{a}=\frac{\text{x}^2-2\text{y}^2}{\text{x}}\ ...(3)$

From (2) and (3), we get

$2\text{x}-\frac{\text{x}^2+2\text{y}^2}{\text{x}}+4\text{y}\frac{\text{dy}}{\text{dx}}=0$

$\Rightarrow2\text{x}^2-\text{x}^2-2\text{y}^2+4\text{xy}\frac{\text{dy}}{\text{dx}}=0$

$\Rightarrow4\text{xy}\frac{\text{dy}}{\text{dx}}+\text{x}^2-2\text{y}^2=0$

$\Rightarrow4\text{xy}\frac{\text{dy}}{\text{dx}}=2\text{y}^2-\text{x}^2$

It is the required differential equation.

Solution of the equaion is given by,